Answer:
characteristics
Explanation:
oranisms are grouped based on shared characteristics
Taxonomists today tend to group organisms into the same category based on their evolutionary history and shared ancestry, as explained in Option A, as this allows taxonomists to construct more accurate and detailed phylogenetic trees.
What is taxonomy?The field of taxonomy is concerned with the classification of living organisms based on their shared characteristics, the aim of taxonomy is to group organisms in a way that reflects their evolutionary history and to provide a standardized system of naming, while earlier taxonomists relied on morphological characteristics, such as the physical appearance of an organism, but modern taxonomists instead use molecular and genetic data to group organisms into categories.
Hence, taxonomists today tend to group organisms into the same category based on their evolutionary histories and shared ancestry, as explained in Option A.
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The question is incomplete, complete question is below
Taxonomists today tend to group organisms into the same category based on similar what?
A)based on their evolutionary histories and shared ancestry
B)based on fossils
If the two chromosomes of a homologous pair carry the same allele for a particular gene, the individual is said to be:_____.
Answer
H0m0zygous
Explanation
The answer to the question is in the attachment. Please open the attachment
For this task, you will imagine that you are a reporter for a scientific magazine. Your task is to explain the process of protein synthesis to someone who does NOT have a science background. Therefore, the explanation needs to be in simple enough terms for anyone to understand. You will organize your article in the following way: Structure and Function of DNA and RNA Transcription and RNA processing Translation Protein modification (general) You must also include the following terms: Double helix Helicase Codon Polymerase 5’ cap Poly (A) tail Introns Exons Splicesomes rRNA, tRNA, Mrna Ribosomes Anticodons E site, P site, A site Initiation Elongation Termination
Answer:
Please find the explanation below
Explanation:
Our nucleus is made up of a genetic material called deoxyribonucleic acid (DNA), which is a double-helical structure that stores the genetic information needed for the optimal functioning of any organism. DNA, alongside RNA are nucleic acids that are composed of NUCLEOTIDES subunits. The nucleotide consists of a pentose sugar (deoxyribose in DNA and ribose in RNA), nitrogenous base and a phosphate group.
However, the genetic information stored in the DNA molecule needs to be expressed in order to form useful products (proteins). This genetic expression is done in two stages viz: transcription and translation. Transcription, which is catalyzed by an enzyme called RNA polymerase is the process whereby the information stored in the DNA is used to synthesize a mRNA molecule. However, this mRNA molecule is considered pre-mature until it is processed. RNA processing occurs in three stages viz: 5' capping, polyadenylation, and splicing.
5' capping involves adding a 5' cap to the marks molecule. Polyadenylation involves adding a poly(A) tail to the mRNA molecule while splicing is the removal of introns (non coding regions) with the aid of Spliceosomes and joining of the exons (coding region). After processing, the mRNA becomes matured and ready to be translated.
Translation is the process whereby the mRNA transcript is used to synthesize a protein molecule. It occurs in the ribosomes (organelles for protein synthesis, a complex of rRNA and proteins) where the mRNA is read in a group of three nucleotides called CODON. The reading is done by the Anticodon of a transfer RNA (tRNA), which is complementary to the codon.
Translation occurs in three stages: initiation, elongation and termination. The mRNA attaches to the P site of the ribosomes (initiation) where it is attached to by a tRNA's anticodon complementary to it. The anticodon carries the amino acid corresponding the codon and shifts to the A-site. The addition of amino acid to the polypeptide chain continues (elongation) until a stop codon is encountered, which signals the end of the translation process i.e. termination. This causes the polypeptide (protein) to be released from the E-site.
The synthesized protein undergoes packaging and modification in the Golgi apparatus.
If a person had not drunk any water for 8 hours before collecting a sample, would you expect his/her urine to have a high or low specific gravity? Why?
Answer:
The person’s urine will have a high specific gravity
Explanation:
A high specific gravity is characterized by an increased concentration of solutes in the urine of am individual. A low specific gravity is characterized by a low or decreased concentration of the solutes.This is usually due to dehydration as a result of environmental conditions or tedious physical activities.
This explains why a person who had not drank any water for 8 hours and is dehydrated before collecting a sample is thought to have a urine with a high specific gravity
Which of the following correctly describes the process of Translation? I. tRNA anticodon bonds to mRNA codon II. Ribosome bonds to mRNA strand III. Ribosome reaches a STOP codon and detaches from the mRNA IV. Each tRNA adds an Amino Acid to the chain as the Ribosome moves along the mRNA V. Complimentary mRNA strand is made from DNA template
Ribosome bonds to mRNA, a tRNA anticodon bonds to mRNA codon, tRNA add an Amino Acid and ribosome detaches from the mRNA describe translation (II, I, IV and III).
What is translation?The translation is the cellular process by which cells can produce proteins by using the genetic code.
The translation involves the use of an mRNA sequence which contains codons that serve to add amino acids to the polypeptide chain in the ribosome.
In conclusion, the steps II, I, IV and III describe translation.
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A student hypothesized that EC growth might be affected by the DNA from circulating erythrocytes. Is this student’s hypothesis reasonable?
Answer:
Explanation:
It can be said that this hypothesis is actually not reasonable because circulating erythrocytes do not contain DNA. This can be proven by the Mammalian erythrocytes (rbcs) which ultimately loose their nuclei as well as many other organelles during their maturation process and they don't have nuclei or mitochondria meaning that they are not affected by the DNA.
different strains of a virus with a segmented genome can shuffle and combine genes.
true
false
Answer:
false
Explanation:
related strains of virus with the signal genome can shuffle and combined genes
Have you ever noticed that aside from zoos, you don't see kangaroos, wombats, and koala bears outside of Australia? This is because there is a trench that is a "biogeographical transition zone between Asian and Australian flora and fauna". What is the name of that trench?
Answer:
Wallace Line
Explanation:
Wallace Line is a boundary line which is a transitional zone between Asia and Australia. This boundary drawn by the British naturalist Alfred Russel Wallace in 1859. This boundary restricts the transferring organisms such as flora and fauna from Australia to Asia. So that's why kangaroos, wombats, and koala bears are only present in Australia not in the continent of Asia.
g Cell autonomous genes affect only those cells in which they are expressed. Which group of early developmental genes are not cell autonomous?
Answer:
The options are not given..
Here are the options.
A. segment polarity genes
B. gap genes
C. coordinate genes
D. homeotic genes
The correct option is A.
segment polarity genes
Explanation:
Segment polarity genes is a group of early developmental genes that are not cell autonomous because their expression helps to define anterior and posterior polarities within each embryonic parasegment. They are the genes that help to function for the polarity within each parasegment of the embryo
Identical twins are formed from one fertilized egg, but fraternal twins form from two different fertilized eggs. Apply your knowledge of mitosis and meiosis to explain the genetic differences between identical and fraternal twins.
Answer:
Identical twins share the exact same DNA while fraternal twins do not.
Explanation:
Mitosis is the type of cell division that results in 2 individual cells that have the exact genetic material.
Meiosis is the division of reproductive cells, eggs and sperms, which results in new cells with half the amount of genetic material and that makes them unique.
Identical twins are twins developed from a single egg that has been fertilized with a single sperm, meaning they have the same DNA.
Fraternal twins are twins from two different eggs that have been fertilized by two different sperms and grow in the womb at the same time with no additional cell division. So they have different DNA's.
I hope this answer helps.
The formation of fraternal twins takes place when the female generates two eggs and identical twins takes place when the female generates one egg and the zygote divides into two spontaneously.
• In identical twins case, the zygote goes through cleavage, that is, cell division by mitosis, when embryo is at the stage of two cell.
• The separation of two blastomeres may take place and then each act as an independent zygote.
• This each dissociated blastomere give rise to a novel independent individual.
• The production of both the organisms takes place due to mitosis, thus, both would possess identical genetic makeup and are known as identical twins.
• Identical twins would be either males or would be either females.
• In case of fraternal twins, the female generates more than one eggs. And all the eggs get fertilized independently by the sperm.
• Thus, formation of two or more zygotes takes place, and each zygote give rise to new offspring.
• These offsprings possess distinct genetic combinations, and such offspring are known as fraternal twins.
• They are produced sexually, and fraternal twins can possess different genders as they do not possess similar genetic recombination.
Thus, identical twins is formed from one egg and the zygote gets divide into two, while fraternal twins formation takes place when two eggs are generated by the female.
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The lemming is a small mammals with a very high daily energy expenditure for its size. Based on what you learned in the panda case study, what would you predict about the lemming
Answer is give below
Explanation:
In the panda case study, we know that here lemming is a small mammal with the very high daily energy expenditure for its size so as per this predict about the lemming that there would not be an affect since the human body does not need any fat and the small intestine would not be able to absorb fats.
What substance heats up the fastest: water, dry sand, wet sand, rock
Answer:
The specific heat capacity represents the amount of energy, in joules, that it takes to raise the temperature of one gram of a given substance by one degree Celsius. Put more simply, the amount of energy it takes to raise a quantity of water by one degree Celsius would raise an equivalent quantity of sand by a little over 14 degrees. Likewise, sand does not need to lose nearly as much energy as water to produce equivalent cooling. Since it "holds" a lot less energy, it cools down much faster than sand.
Indeed, liquid water has an unusually high specific heat capacity. Because it is much less prone to temperature swings than other common substances, large bodies of water often work to moderate temperatures in a region. This helps to explain, for example, why average temperatures fluctuate very little over the year in San Francisco, a cit
E
A class is debating pros and cons of using sea walls to reduce beach erosion. What is the best way for them inform their
decision-making?
They should each write an opinion paragraph to send to the local paper.
They should hold the debate on stage, so that they can be judged.
They should watch movies about beaches.
They should research wave action and collect data about erosion in the area.
Answer:
They should research wave action and collect data about erosion in the area.
Explanation:
everyone has their own opinion, and movies aren't gonna make the cut, and no one wants to see a debate in school, so they should research first to get the info they need
Two scientists did the same experiment but arrived at different results. The scientists most likely
Answer:
Two scientists did the same experiment but arrived at different results. The results would most likely provide new testable ideas.
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Answer:
If two scientists did the same experiment but arrived at different results it would most likely provide new testable ideas.
1. Research 5 functions of the frontal lobe and explain them in detail with example.
2. Explain what you think might happen if someone damaged frontal lobe. You must mention the functions you have listed
Answer:
the front lobes are involved in motor function, problem solving,spontaneity,memory,language,initiation,judgement,impulse control,and social behaviour..
Explanation: particulary, an injury of the frontal lobecould leads to deficits in executive function such as anticipation , goal selection, planning and self correction
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in the krebs cycle,
a. pyruvate is broken down into Co2 molecules
b. glucose is broken down into NADH and CO2
c. Co2 is concerted in pyruvate and O2
d. NADH and CO2 are converted into ATP
Answer:
a. pyruvate is broken down into Co2 molecules
Explanation:
Krebs cycle is also known as Citric Acid Cycle, in which pyruvate formed during the glycolytic is broken down into Carbon Dioxide (CO2) and Water (H2O).
Each pyruvate molecule is broken down into carbon atom which releases carbon dioxide.
Hence, the correct answer is "a. pyruvate is broken down into Co2 molecules".
Carbon first enters the food web in Multiple Choice humans. vertebrates. invertebrates. green plants. soil bacteria
Answer:
Green Plants
Explanation:
Carbon enters all food webs, both terrestrial and aquatic, through autotrophs, or self-feeders.
Which of the circumstances below most accurately describes conditions that are likely to permit a robust anti-tumor adaptive immune response?
a. A tumor with high expression of proteins that have mutations in sequences encoding HLA-binding peptides.
b. A robust acute inflammatory response to PAMPs expressed specifically by tumor cells.
c. Anti-tumor T lymphocytes that have high expression of CTLA-4.
d. A tumor with high expression of PD-L1.
Did you ever figure out the answer to this question?
The study of PAMP-DAMP complexes is vital to the advancement of knowledge regarding inflammatory disorders in general and cancer in particular. Therefore, option (B) is correct.
What is PAMP-DAMP complexes?Increasing evidence links inflammation to cancer, and at the root of inflammation are PAMPs and DAMPs (DAMPs). Microorganisms contain PAMPs, which are detected by pattern recognition receptors on monocytes and DCs (PRRs). PRR activation produces pro-inflammatory cytokines.
A robust immune response requires endogenous chemicals that pose 'risk' to self-tissues and are created by injured or stressed cells; these are DAMPs, which also trigger inflammation. PAMPs and DAMPs each have about 100 receptors. PAMPs and DAMPs interact; a PRR can bind to both. In this system, PAMPs and DAMPs affect each other's activation threshold. Thus, PAMP-DAMP relationships describe inflammation in a predictable 'inflammatory code'
PAMP-DAMP complexes are key to understanding inflammatory disorders and cancer.
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In the image, the arrow Is pointing to a celestial object. Which attribute disqualifies the object from being a planet?
A. It appears to have no moons.
B. It is spherical in shape.
C. Its neighborhood is not clear.
D. It has horizontal bands on its surface.
The correct answer is C. Its neighborhood is not clear.
Explanation:
One of the key features of a planet is that it clears its neighborhood; this means the planet gravitational force is strong enough that there are no other celestial bodies on its orbit or next to it except by its moons or satellites. This does not occur in the image presented because there are smaller celestial bodies near to the main or bigger celestial body. This likely occurs if the celestial body is not big enough to clear its neighborhood, and therefore it is not a planet.
Galactosemia is a metabolic disorder characterized by the inability to metabolize the sugar galactose. People with galactosemia suffer from liver, kidney, and brain damage among other symptoms. A gene mutation underlying galactosemia was identified by sequencing the genome of a person with galactosemia. What additional evidence would support the hypothesis that the mutation in the candidate gene causes the disease
Answer:
Genetic mapping for unequivocal identification of the potentially causative mutation
Explanation:
Galactosemia is a genetic disorder caused by mutations in the Galactose-1-phosphate uridylyltransferase (GALT) gene, which encodes an enzyme involved in the metabolism of galactose. Gene mapping is a technique widely used in genetics to identify the position of one locus a chromosome by using molecular markers to estimate genetic distances. Genetic mapping provides useful evidence in order to identify when a disease that is transmitted from parent to offspring can be associated with one or more genes and then determine which gene/s is/are responsible for this condition.
Is zinc renewable or not?
Answer: Zinc is non renewable.
Explanation:
Non-Renewable resources are resources that cannot be replenished easily through natural process. These are the sources which are used in wise manner so that they do not get depleted. For example: coal, petroleum etc.
Renewable resources are resources that can be replenished overtime through natural process.These are the sources which never depletes completely. For example: bio mass energy, solar energy etc.
As zinc cannot be replenished easily through natural process, it is non renewable.
Squamous cells specialize in what function? A) the secretion of enzymes B) diffusion of gasses C) the absorption of nutrients D) none of the above
Answer:
The abosbton of nutrients
The ____ ring is built onto ribose-5-phosphate of PRPP for its de-novo nucleotide biosynthesis, while the ring structure of the ______ bases are synthesized separately and then coupled to ribose-5-phosphate via the C-N glycosidic bond.
Answer:
The purine ring is built onto ribose-5-phosphate of PRPP for its de-novo nucleotide biosynthesis, while the ring structure of the pyrimidine bases are synthesized separately and then coupled to ribose-5-phosphate via the C-N glycosidic bond.
Explanation:
In the de novo synthesis of nucleotides, their metabolic precursors such as aminoacids, ribose-5-phosphate, CO₂ and NH₃ are used as starting materials.
In purine nucleotide synthesis, the ring structure is built up on ribose-5-phosphate of PRPP by addition of one or a few atoms one at a time starting with the amino group donated by glutamine until the first intermediate inosinate is synthesized.
In pyrimidine ring synthesis, orotate is first synthesized from carbamoyl phosphate and aspartate, and then is attached to ribose-5-phosphate of PRPP, before it is then converted to the common pyrimidine nucleotides starting from uridylate.
You have been assigned a DNA stretch. What is the complementary strand when you replicate the template
Answer:
The complementary strands of a DNA are read in the opposite direction to one another
Explanation:
A DNA strand/stretch can be represented with a number of bases. The bases in a DNA strand are Adenine (A), Thymine (T), Cytosine (C) and Guanine (G). The Adenine can bind to Thymine (and vice-versa) while the Guanine can bind to Cytosine (and vice versa). These bonded bases are called "base pairs".
Complementary DNA strands are read in opposite direction to one another. Each end of a DNA strand is either a 5' end or a 3' end. Hence, if one strand runs from the 5' to 3' end, the complementary strand will run from the 3' to 5'. Thus, they are represented "on paper" the way they run.
For example,
If the DNA strand assigned is AGTCTAG running 5' to 3', the complementary strand will be CTAGACT (starting from the 3' end) and NOT TCAGATC (which starts from the 5' end)
As part of your research, you examine mRNA expression for the parents, their two unaffected children, and their affected child with a Northern blot using a full-length cDNA probe.Your results, shown here, are as follows:Both parents have low, but detectable, mRNA expression.The unaffected children have either normal or low mRNA expression.The affected child has no detectable mRNA expression. What possible molecular defect could explain these results?Select every legitimate possibility from the choices below.a. The recessive allele could be a premature stop codon that prevents the translation of a full-length protein.b. The recessive allele could be a mutation in the promoter region that prevents RNA polymerase II from binding.c. The recessive allele could be a mutation in the transcribed region that makes the mRNA unstable and causes it to degrade.d. The recessive allele could be a mutation in regulatory DNA that prevents chromatin from being remodeled from a closed to an open state.e. The recessive allele could be a mutation in a splice donor or acceptor sequence that prevents proper mRNA splicing, leaving an intron in the spliced transcript.
Answer:
b.c d are the answer options
Explanation:
b. The recessive allele could be a mutation in the promoter region that prevents RNA polymerase II from binding.
c. The recessive allele could be a mutation in the transcribed region that makes the mRNA unstable and causes it to degrade.
d. The recessive allele could be a mutation in regulatory DNA that prevents chromatin from being remodeled from a closed to an open state.
in option b, the synthesis of mRNA is going to be halted
in option c, mRNA is undergoing degradation, it will be invisible in the northern blot.
in option d, mRNA in expression of each gene is going to be turned off.
.
In quiet breathing, muscular effort is used mainly in inspiration, and expiration is largely passive, due to elastic recoil of the lung. It results in a pattern change between expiratory and inspiratory flow. Explain changes you observed in the recordings you obtained.
Answer:
The summary of something like the particular question is demonstrated in the following portion on either the clarity.
Explanation:
Inspiration depends on either the diaphragm contracting, as well as the influence including its external intercostal has been limited throughout normal breathing. Maximum inhalation or even just expiration time depth, as well as rate, are fairly directly proportional to one another because.We may link the whole observation to something like the sequence of expiratory versus inspiratory movement as it takes much longer than either the inspiratory speed throughout our sample since expiration becomes passive yet requires having to return to something like the steady levels.In angiosperms, what is the name of the outer epidermal layer that protects the plant body?
Answer:
cuticle
Explanation:
To study the fate of carbon during glycolysis under aerobic conditions, cultured cells are fed with 14C-glucose labeled in various positions. You use a radiolabeled form of glucose where the radioactive 14C occupies position 1. Which of the following is correct?a. The carboxyl group of acetyl-CoA and CO2 become radiolabeledb. The methyl group of acetyl CoA and CO2 become radiolabeledc. The carbonyl group of acetyl-CoA becomes radiolabeledd. The methyl group of acetyl CoA becomes radiolabelede. CO2 becomes radiolabeledf. I don’t have enough information to conclude
Answer:
D. The methyl group of acetyl CoA becomes radio-labeled
Explanation:
During the steps in glycolysis, the carbon at position 1, becomes C-1 in dihydroxyacetone phosphate during the cleavage of fructose-1,6-bisphosphate to dihydroxyacetone phosphate and glyceraldehyde-3-phosphate. Subsequently on isomerization of dihydroxyacetone phosphate to glyceraldehyde-3-phosphate, C-1 of dihydroxyacetone phosphate becomes C-3 of glyceraldehyde-3-phosphate.
Furthermore, in pyruvate, the end product of glycolysis, C-3 is converted to a methyl group which then becomes the methyl group in the acetyl-CoA molecule produced from the oxidative decarboxylation of pyruvate.
Since the radioactive 14-C of radio-labeled glucose occupies position 1, it will become the methyl group of acetyl-CoA.
A container on the counter labeled "ground beel" is not properly date marked. The cook tells the Person in charge (PIC) that it was
prepared recently. The Person in charge (PIC) should
Ouke the ground beer in temperature is alr(**) or lower
O ada the ground beer to chilor Noup.
O discard the ground beer
O use the ground beef on pizza prepared for immediate service.
Answer:
Correct Answer:
O discard the ground beer
Explanation:
When it comes to anything edible whether food or drink, safety and precaution is always taken. This is to prevent poisoning when an expired or harmful food or drink is taken. In the case of the undated "ground beer" on the counter, the best action would be to discard it entirely.
In the given case, The Person in charge (PIC) should - discard the ground beer
All foods that require time and temperature control (TCS) should be labeled with the following:
Common name of the food Date the food was made.Use by date.Date marking is a process assuring the food is discarded before these bacteria can cause foodborne illness.anything edible whether food or drink, safety and precaution is always taken.If something not marked properly should be discardedprevent poisoning when an expired or harmful food or drink is taken.Thus, In the case of the undated "ground beer" on the counter, the best action would be to discard it entirely.
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You decide to focus first on Mutation 1 and Mutation 2.
The F' plasmid appears to "rescue" mutant strain 1 by restoring wild-type inducible expression.
The F' plasmid does not change the phenotype of mutant strain 2; there is still no expression of the lac operon.
There are three possible mutations that could lead to lack of expression: a super-repressor (Is), a promoter mutation (P-) or a mutation in the lacZ gene (Z- ). To determine the effect of introducing a wild-type F' plasmid into a bacterial cell with each of these mutations, you draw out the three possible scenarios.
F' l+ cap+
Determine the expression of the lac operon for each possible partial diploid and for the "unknown" mutation partial diploids (Mutation 1 and Mutation 2).
1. F' lac+/ ts
2. F' lac+/ P
3. F' lac+/ Z
4. F' lac+/ Mutation 1
5. F lac+/ Mutation 2
A. Inducible (wild type)
B. No expression (mutant)
Answer:
Super-repressor (Is) - Uninducible dominant
Binding region is modified by the protein called repressor which prevents the lactose to bind with it. Thus, even the lactose is present the repressor protein is bound to the promoter and can not be detach from it. Transcription is not inducible.
Promoter mutation (P)- - Uninducible recessive
Mutated promoter region inhibits the binding of RNA polymerase that leads to trranscription.
The lacZ gene ( Z-) is mutated - recessive while the normal LacZ gene (Z +) - dominant .
Inducible (Wild type) No expression (Mutant)
F' lac+/ Is No Yes
F' lac + / P- Yes No
F' lac + / Z- Yes No
F' lac +/ Mutation1 Yes No
F' lac+ / Mutation2 No Yes
Mutation 1 and F' plasmid - Inducible wild type
Mutation 2 and F' plasmid - Mutant no expression
function of carbohydrates
Answer:
Alongside fat and protein, carbohydrates are one of the three macronutrients in our diet with their main function being to provide energy to the body. They occur in many different forms, like sugars and dietary fiber, and in many different foods, such as whole grains, fruit and vegetables.
Explanation:
